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. South Carolina State Uni x S. www.sap inglearning.com × . www.saplinglearning.com Li www.saplinglearning.com/ib.scms/mod/ibis/view.php?id=2620878 6/15/2016 11:55 PM A 0/100 6/15/2016 07:09 PM Assignment Information Available From Not Set Due Date: Late submissions: Allowed with 3% of the # Attempts Gradebook Print Calculator Periadic Table Question 8 of 10 6/15/2016 11:55 PM Mapoob Sapling Learning macmilan learning points possible deducted per day until 06/20/2016 11:55 PM A particle with mass 2.57 kg oscillates horizontally at the end of a horizontal s amplitude of 0.843 m and a duration of 121 s for 75 cycles of oscillation. Find the frequency, f, the speed at the equilibrium position, Vmax, the spring constant, k, the potential energy at an endpoint, Umax, the potential energy when the particle is located 51.1% of the amplitude away from the equilibrium position, U, and the kinetic energy, K, and the speed, v, at the same position A student measures an 4 Points Possible Grade Category: Description: Policies: 100 Graded Number new-hw You can check your answer You can view solutions after the due date You have five attempts per question. There is no penalty for incorrect answers. s. m/s 10 Number N/ m eTextbook Number (Scroll down for more answer blanks.) D Help With This Topic Web Help & Videos Previous O Check Answer Next Exit Technical Support and Bug Reports Hint © 2011-2016 Sapling Learning, Inc.-139 about us careers partners privacy policy terms of use contact ushelp 8:12 PM 6/15/2016Explanation / Answer
amplitude=0.843 m
frequency= 75/121= 0.619834711 Hz
w= 2pi f =3.89256
X= Asinwt = 0.843sin(3.89256t)
V= 3.28142975342cos(3.89256t)
So, Vmax= 3.2814 m/s
w=(k/m)0.5
=> k= 38.9407 N/m
Umax= KEmax = 0.5*2.57*3.2814*3.2814= 13.8363 J
U = 0.5kx2
At 51.1% amp, U= 0.5*38.9407*(0.511*0.843*0.511*0.843)= 3.6130 J
KE= TE-PE= 10.2232 J
=>V=(KE/(0.5m))0.5 = 2.8206 m/s
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